Jake ate 1/6 of the chocolates the jill came ate 1/5 of the remaining then mike ate 1/4 of the remaining then sara ate 1/3 of the remaining and luke ate 1/2 of what was left. i came home and there was 4 chocolate left. how many did jake eat?

  4 chocolates

Step-by-step explanation:

Before Luke started eating, there were 8. This is 2/3 of what Sara had to start with, so she started with 12. That is 3/4 of what Mike started with, so he started with 16.

Jill left 4/5 of what she started with, so she must have started with 20. Jake ate 1/6 and left 5/6, so 20 is 5 times the number Jake ate.

Jake ate 4 chocolates.

___

Jill ate 4; Mike ate 4; Sara ate 4; and Luke ate 4.

option a, c,d are correct.

step-by-step explanation:

from the given figure, it is given that z is equidistant from the sides of the triangle rst, then from triangle tzb and triangle szb, we have

tz=sz(given)

bz=zb(common)

therefore, by rhs rule,δtzb ≅δszb

by cpctc, sz≅tz

also, from δctz and δasz,

tz=sz(given)

∠tcz=∠saz(90°)

by rhs rule, δctz ≅ δasz, therefore by cpctc, ∠ctz≅∠asz

also,from δasz and δzsb,

zs=sz(common)

∠zbs=∠saz=90°

by rhs rule, δasz ≅δzsb, therefore, by cpctc, ∠asz≅∠zsb

hence, option a, c,d are correct.